Write and explain sum of subset algorithm for W = {2, 7, 8, 9, 15} M = 17

1. Given positive numbers (weight) w_i where (1<=i<=n) and m.
2. This problem calls for finding all subsets of w_i whose sum is m.
3. For e.g. if n=4, m=31, (w_1,w_2,w_3,w_4) = (11, 13, 24, 7).
4. Then the desired subset are (11, 13, 7) and (2, 4, 7).
5. In this method of giving the solution, the solution vector is given as variable-sized tuple (tuple is collection of element).

6. The solution may also be expressed as fixed size tuple (x_1,x_(2,),x_3,…….,x_n) such that

   x_i =0 if w_i is not selected

   x_i =1 if w_i is selected

7. Using this strategy the solution would be (1,1,0,1) and (0,0,1,1)

Algorithm:

• Let n be the number of elements and let m be the required sum.
• Let w[1…..n] be an array of weights in ascending order and Let x[1….n] be the solution vector.

• Method

  SumOfSubset(s, k, r)
  
  {
  x[k] =1;     //Generating left child
  if(s+w[k]==m)
      Display x[1….k]  //and remaining zeros
  else if(s+w[k]+w[k+1]<=m)
      SumOfSubset(s+w[k], k+1, r-w[k])
  if((s+r-w[k]>=m) && (s+w[k+1]<=m))
  {
      x[k]=0;
      SumOfSubset(s, k+1, r-w[k]);
  }
  }
  

• S → existing sum

• w[k] → weight of current element
• w[k+1] → weight of next element
• s+w[k]+w[k+1]<=m → sum of existing element+weight of current element+weight of next element is less tham m.
• Thus the next element can be selected for time being.
• s+r-w[k]>=m → Even if u discard the current element, you may still get the required sum later on.
• s+w[k+1]<=m → The sum of selected element + weight of next element is less than m . This is even if u take the next element the sum will not exit the required sum(m)
• Thus next element can be selected.

Problem:

n = 5, W = {2, 7, 8, 9, 15} M = 17

Hence, there are three subset with given sum m=17

A(111)

B(10001)

C(0010)